A000140 Kendall-Mann numbers: the most common number of inversions in a permutation on n letters is floor(n*(n-1)/4); a(n) is the number of permutations with this many inversions.

1, 1, 2, 6, 22, 101, 573, 3836, 29228, 250749, 2409581, 25598186, 296643390, 3727542188, 50626553988, 738680521142, 11501573822788, 190418421447330, 3344822488498265, 62119523114983224, 1214967840930909302, 24965661442811799655, 538134522243713149122

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• core
• nice

Comments

Row maxima of A008302, see example.

The term a(0) would be 1: the empty product is one and there is just one coefficient 1=x^0, corresponding to the 1 empty permutation (which has 0 inversions).

From Ryen Lapham and Anant Godbole, Dec 12 2006: (Start)

Also, the number of permutations on {1,2,...,n} for which the number A of monotone increasing subsequences of length 2 and the number D of monotone decreasing 2-subsequences are as close to each other as possible, i.e., 0 or 1. We call such permutations 2-balanced.

If 4|n(n-1) then (with A and D as above) the feasible values of A-D are C(n,2), C(n,2)-2,...,2,0,-2,...,-C(n,2), whereas if 4 does not divide n(n-1), A-D may equal C(n,2), C(n,2)-2,...,1,-1,...,-C(n,2). Let a_n(i) equal the number of permutations with A-D the i-th highest feasible value.

The sequence in question gives the number of permutations for which A-D=0 or A-D=1, i.e., it equals A_n(j) where j = floor((binomial(n,2)+2)/2). Here is the recursion: a_n(i) = a_n(i-1) + a_{n-1}(i) for 1 <= i <= n and a_n(n+k) = a_n(n+k-1) + a_{n-1}(n+k) - a_n(k) for k >= 1. (End)

The only two primes found < 301 are for n = 3 and 6.

Define an ordered list to have n terms with terms t(k) for k=1..n. Specify that t(k) ranges from 1 to k, hence the third term t(3) can be 1, 2, or 3. Find all sums of the terms for all n! allowable arrangements to obtain a maximum sum for the greatest number of arrangements. This number is a(n). For n=4, the maximum sum 7 appears in 6 arrangements: 1114, 1123, 1213, 1222, 1231, and 1132. - J. M. Bergot, May 14 2015

Named after the British statistician Maurice George Kendall (1907-1983) and the Austrian-American mathematician Henry Berthold Mann (1905-2000). - Amiram Eldar, Apr 07 2023

Examples

			From _Joerg Arndt_, Jan 16 2011: (Start)
a(4) = 6 because the among the permutations of 4 elements those with 3 inversions are the most frequent and appear 6 times:
       [inv. table]  [permutation]  number of inversions
   0:    [ 0 0 0 ]    [ 0 1 2 3 ]    0
   1:    [ 1 0 0 ]    [ 1 0 2 3 ]    1
   2:    [ 0 1 0 ]    [ 0 2 1 3 ]    1
   3:    [ 1 1 0 ]    [ 2 0 1 3 ]    2
   4:    [ 0 2 0 ]    [ 1 2 0 3 ]    2
   5:    [ 1 2 0 ]    [ 2 1 0 3 ]    3  (*)
   6:    [ 0 0 1 ]    [ 0 1 3 2 ]    1
   7:    [ 1 0 1 ]    [ 1 0 3 2 ]    2
   8:    [ 0 1 1 ]    [ 0 3 1 2 ]    2
   9:    [ 1 1 1 ]    [ 3 0 1 2 ]    3  (*)
  10:    [ 0 2 1 ]    [ 1 3 0 2 ]    3  (*)
  11:    [ 1 2 1 ]    [ 3 1 0 2 ]    4
  12:    [ 0 0 2 ]    [ 0 2 3 1 ]    2
  13:    [ 1 0 2 ]    [ 2 0 3 1 ]    3  (*)
  14:    [ 0 1 2 ]    [ 0 3 2 1 ]    3  (*)
  15:    [ 1 1 2 ]    [ 3 0 2 1 ]    4
  16:    [ 0 2 2 ]    [ 2 3 0 1 ]    4
  17:    [ 1 2 2 ]    [ 3 2 0 1 ]    5
  18:    [ 0 0 3 ]    [ 1 2 3 0 ]    3  (*)
  19:    [ 1 0 3 ]    [ 2 1 3 0 ]    4
  20:    [ 0 1 3 ]    [ 1 3 2 0 ]    4
  21:    [ 1 1 3 ]    [ 3 1 2 0 ]    5
  22:    [ 0 2 3 ]    [ 2 3 1 0 ]    5
  23:    [ 1 2 3 ]    [ 3 2 1 0 ]    6
The statistics are reflected by the coefficients of the polynomial
(1+x)*(1+x+x^2)*(1+x+x^2+x^3) ==
x^6 + 3*x^5 + 5*x^4 + 6*x^3 + 5*x^2 + 3*x^1 + 1*x^0
There is 1 permutation (the identity) with 0 inversions,
3 permutations with 1 inversion, 5 with 2 inversions,
6 with 3 inversions (the most frequent, marked with (*) ), 5 with 4 inversions, 3 with 5 inversions, and one with 6 inversions. (End)
G.f. = x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 101*x^6 + 573*x^7 + 3836*x^8 + ...
		

References

• F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 241.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Robert Israel, Robert G. Wilson v and N. J. A. Sloane Table of n, a(n) for n = 1..400 (a(1) to a(61) from Sloane, a(62) to a(350) from Wilson).
• Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
• Dominique Foata, Distributions eulériennes et mahoniennes sur le groupe des permutations, pp. 27-49 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
• Mikhail Gaichenkov, The property of Kendall-Mann numbers, MathOverflow, 2010.
• Richard K. Guy, Letter to N. J. A. Sloane with attachment, Mar 1988.
• A. Waksman, On the complexity of inversions, IEEE Trans. Computers, 19 (1970), 1225-1226.
• Wikipedia, Inversion.
• Wikipedia, q-Pochhammer symbol. - _Paul Muljadi_, Jan 18 2011
• Index entries for "core" sequences.

Crossrefs

Row maxima of A008302.

Odd terms are A186888.

Cf. A000124, A001147, A128566, A181609.

Programs

Magma

/* based on David W. Wilson's formula */ PS:=PowerSeriesRing(Integers()); [ Max(Coefficients(&*[&+[ x^i: i in [0..j] ]: j in [0..n-1] ])): n in [1..21] ]; // Klaus Brockhaus, Jan 18 2011

Maple

f := 1: for n from 0 to 40 do f := f*add(x^i, i=0..n): s := series(f, x, n*(n+1)/2+1): m := max(coeff(s, x, j) $ j=0..n*(n+1)/2): printf(`%d,`,m) od: # James Sellers, Dec 07 2000 [offset is off by 1 - N. J. A. Sloane, May 23 2006]
P:= [1]: a[1]:= 1:
for n from 2 to 100 do
P:= expand(P * add(x^j,j=0..n-1));
a[n]:= max(eval(convert(P,list),x=1));
od:
seq(a[i],i=1..100); # Robert Israel, Dec 14 2014

Mathematica

f[n_] := Max@ CoefficientList[ Expand@ Product[ Sum[x^i, {i, 0, j}], {j, n-1}], x]; Array[f, 20]
Flatten[{1, 1, Table[Coefficient[Expand[Product[Sum[x^k, {k, 0, m-1}], {m, 1, n}]], x^Floor[n*(n-1)/4]], {n, 3, 20}]}] (* Vaclav Kotesovec, May 13 2016 *)
Table[SeriesCoefficient[QPochhammer[x, x, n]/(1-x)^n, {x, 0, Floor[n*(n-1)/4]}], {n, 1, 20}] (* Vaclav Kotesovec, May 13 2016 *)

PARI

{a(n) = if( n<0, 0, vecmax( Vec( prod(k=1, n, 1 - x^k) / (1 - x)^n)))}; /* Michael Somos, Apr 21 2014 */

Python

from math import prod
from sympy import Poly
from sympy.abc import x
def A000140(n): return 1 if n == 1 else max(Poly(prod(sum(x**i for i in range(j+1)) for j in range(n))).all_coeffs()) # Chai Wah Wu, Feb 02 2022

Formula

Largest coefficient of (1)(x+1)(x^2+x+1)...(x^(n-1) + ... + x + 1). - David W. Wilson

The number of terms is given in A000124.

a(n+1)/a(n) = n - 1/2 + O(1/n^{1-epsilon}) as n --> infinity (compare with A008302, A181609, A001147). - Mikhail Gaichenkov, Apr 11 2014

Asymptotics (Mikhail Gaichenkov, 2010): a(n) ~ 6 * n^(n-1) / exp(n). - Vaclav Kotesovec, May 17 2015

Extensions

Edited by N. J. A. Sloane, Mar 05 2011